Problem 7: A 4 kg traveling with a speed of 2 m/s strike a rigid wall and rebounds elastically. If the ball is in contact with the wall for 0.05 sec, what is

• the momentum imparted to the wall
• the average force exerted on the wall

Solution

Given: Mass of the ball, m = 4 kg,              v_i = 2 m/s,            v_f = -2 m/s,          Δt = 0.05 sec

Find: Momentum imparted (passed on) to the wall and the average force on the wall.

Note

• The collision is elastic, therefore, net change in the momentum will be zero.
• If the velocity toward the wall is taken positive, the rebounded velocity will be negative.

 Now initial momentum = p_i = mv_i = 4 × 2 = 8 kg ms^-1

Final momentum =              p_f = mv_f = 4 × (-2) = – 8 kg ms^-1

Therefore, change in momentum of the ball = Δp = p_f – p_i = – 8 – 8 = – 16 kg ms^-1

This change in the momentum of the ball is passed on to the wall so that the change in the momentum of the system is zero.

To find the average force on the wall:

From Newton’s 2^nd law of motion, F = ma = m (v/t). Therefore, momentum expressed as Newton’s 2^nd law is, F× t = mv = p. Therefore, change in momentum = Δp = mv_f – mv_i = m (v_f – v_i). Hence,

F × Δt = m × Δv. Put the values,

F × 0.05 = 2 × [-8 – 8)] = -2 × 16 = -16

OR  F = 16/0.05 = -320 N

Hence, the average force exerted on the wall is 320 N.